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Magazine Chapter 21: Problem 20
If \(\sum_{i=1}^{n} a_{i}=0\) where \(\left|a_{i}\right|=1+i\), then the value of
is \(\sum_{1 \leq i
Short Answer
Expert verified
The correct answer is \(-\frac{n}{2}\), which corresponds to option (A).
Step by step solution
01
Understanding the Problem
We need to compute the double sum \(\sum_{1 \leq i < j \leq n} a_i \cdot a_j\). Given that the total sum of all \(a_i\) is zero, and \(|a_i| = 1 + i\), we are to find the value of this double sum.
02
Analyzing Given Conditions
Each \(a_i\) has a magnitude of \(1 + i\) but can be positive or negative. For the sum of all \(a_i\) to be zero, some \(a_i\) must be positive and some negative. This suggests a symmetric distribution of these values around zero.
03
Rewriting the Double Sum
Note that \(\sum_{i=1}^{n} a_i^2\) is easy to compute as each \(a_i^2 = (1+i)^2\). Use the identity \((\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{1 \leq i < j \leq n} a_i \cdot a_j\) with \(\sum_{i=1}^{n} a_i = 0\) to get \(0 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{1 \leq i < j \leq n} a_i \cdot a_j\).
04
Computing Sum of Squares
Calculate \( \sum_{i=1}^{n} a_i^2 = \sum_{i=1}^{n} (1+i)^2 \). This results in \( \sum_{i=1}^{n} (1+i)^2 = \sum_{i=1}^{n} (1 + 2i + i^2) = \sum_{i=1}^{n} 1 + 2 \sum_{i=1}^{n} i + \sum_{i=1}^{n} i^2 \).
05
Evaluating Arithmetic Series
Use known formulae: \(\sum_{i=1}^{n} i = \frac{n(n+1)}{2}\) and \(\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}\). Substitute these in to get \(n + n(n+1) + \frac{n(n+1)(2n+1)}{6}\).
06
Solving for Double Sum
Substitute back into the identity: \(0 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{1 \leq i < j \leq n} a_i \cdot a_j\). We have the sum of squares from the previous step, solve for \(\sum_{1 \leq i < j \leq n} a_i \cdot a_j = -\frac{1}{2} \sum_{i=1}^{n} a_i^2\).
07
Finding the Final Result
Compute \(-\frac{1}{2}\) times the polynomial expression found for \(\sum_{i=1}^{n} a_i^2\). After these calculations, you find the final simplified result for the problem, yielding \(-\frac{n(n+1)(2n+1) + 6n(n+1) + 6n}{12} = -\frac{n}{2}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Summation Techniques
When we talk about summation techniques in mathematics, we refer to methods for adding up numbers in a series or sequence. For example, techniques such as the arithmetic series and the squared series can simplify complex summations.
In this exercise, we deal with double summations where sums are calculated over a range of indexed terms. Specifically, the exercise involves computing \( \sum_{1 \leq i < j \leq n} a_i \cdot a_j \).
In this exercise, we deal with double summations where sums are calculated over a range of indexed terms. Specifically, the exercise involves computing \( \sum_{1 \leq i < j \leq n} a_i \cdot a_j \).
- Symmetric Sum: This is key because we're told \( \sum_{i=1}^{n} a_i = 0 \). It implies that the values are symmetrically distributed around zero.
- Utilizing Identities: The identity \( (\sum_{i=1}^{n} a_i)^2 = \sum_{i=1}^{n} a_i^2 + 2 \sum_{1 \leq i < j \leq n} a_i \cdot a_j \) facilitates finding the double sum. By knowing that the first sum part becomes zero, it simplifies the equation to solve for the double summation.
Algebraic Manipulation
Algebraic manipulation involves rewriting expressions to simplify calculations or to reveal deeper relationships among terms. This exercise relies heavily on these skills to derive the result.
The given relation for \( a_i \) suggests a clever choice in expressing each term from magnitude and sign conditions to fulfill \( \sum_{i=1}^{n} a_i = 0 \).
The given relation for \( a_i \) suggests a clever choice in expressing each term from magnitude and sign conditions to fulfill \( \sum_{i=1}^{n} a_i = 0 \).
- Breaking Down the Terms: Rewriting \( \sum_{i=1}^{n} (1+i)^2 \) by expanding each term to isolate central components: \( 1 + 2i + i^2 \).
- Using Formulas: Known sums for \( i \) and \( i^2 \) are applied;
- Sum of first \( n \) natural numbers: \( \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \).
- Sum of squares: \( \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \).
Series and Sequences
Series and sequences form a fundamental part of understanding mathematical problems that involve repeated sums. In our problem, we observe the properties of arithmetic series that aid in solving the summation.
We begin with understanding the nature of \( a_i \) values as part of a structured sequence with predictable changes (increments by 1 with each step). The importance of sequences is not only in calculating sums but in recognizing patterns, such as the requirement for symmetry for a net sum of zero.
We begin with understanding the nature of \( a_i \) values as part of a structured sequence with predictable changes (increments by 1 with each step). The importance of sequences is not only in calculating sums but in recognizing patterns, such as the requirement for symmetry for a net sum of zero.
- Arithmetic Series: It occurs when each term increases by a constant value, which allows easy computation of the sum using standard formulas.
- Understanding Sequences: By observing sequences like \( |a_i| = 1+i \), which help frame the symmetry needed for zero summation, we leverage this structured approach for complex problems.
